Chapter 26, Problem 34P

To determine

The rest mass and energy lost in the collision. Also, quantity of lost kinetic energy in this collision.

Answer to Problem 34P

The rest mass: $M_0=\frac{2m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

The energy loss is 0.

Quantity of lost energy: $2m_0{c}^2\left(\left\{\frac{1}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}\right\}-1\right)$

Explanation of Solution

Given:

Potential difference is $25,000V=0.025MV$

The rest mass of an electron is $0.511MeV/{\text{c}}^2$

Formula used:

It is known that,

  $m=\frac{m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

Where,

  $m_0$ is rest mass

  $m$ is relativistic mass

  $v$ is velocity

  $c$ is a speed of light, $3.00\times {10}^{8}m/s$

Calculation:

First calculate the rest mass.

If ${\text{m}}_0$ is the rest mass of the new particle, for conservation of energy then:

  $2m{c}^2=\frac{2m_0{c}^2}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}=M_0{c}^2$

  $M_0=\frac{2m_0}{{\left[1-{\left(\frac{v}{c}\right)}^2\right]}^{\frac{1}{2}}}$

According to this, energy is conserved. So, there was no energy loss.

Also, final particle is at rest. Thus, kinetic energy loss is the initial kinetic energy of the two colliding particles:

  $K{E}_{loss}=2\left(m-m_0\right)c^2=2m_0{c}^2\left(\left\{\frac{1}{{\left[1-\left(\frac{v^2}{c^2}\right)\right]}^{\frac{1}{2}}}\right\}-1\right)$